2sin^2(theta) = 3 - 4cos(theta).
since sin^2(theta) = 1 - cos^2(theta), substitute for sin^2(theta) to get:
2 * (1 - cos^2(theta) = 3 - 4cos(theta)
simplify to get:
2 - 2cos^2(theta) = 3 - 4cos(theta)
add 2cos^2(theta) and subtract 2 from both sides of the equation to get:
0 = 3 - 4cos(theta) + 2cos^2(theta) - 2
combine like terms and flip the equation to get:
2cos^2(theta) - 4cos(theta) + 1 = 0
that's your quadratic equation that you need to factor.
use the quadratic formula to factor it as follows:
since the equation is in standard form of ax^2 + bx + c = 0, you get:
a = 2
b = -4
c = 1
the quadratic formula is:
cos(theta) = -b plus or minus sqrt(b^2 - 4ac)
--------------------------------
2a
replace a,b,c with their values and you get:
cos(theta) = -(-4) plus or minus sqrt((-4)^2 - 4*2*1)
--------------------------------
2*2
you will get:
cos(theta) = (4 + sqrt(8))/4
or:
cos(theta) = (4 - sqrt(8))/4
the result will be:
cos(theta) = 1.707106781
or:
cos(theta) = .2928932188
cosine can't be greater than 1, so the solution is:
cos(theta) = .2928932188
solve for theta to get:
theta = 72.96875154 degrees.
that would be in quadrant 1.
cosine is positive in quadrants 1 and 4 only.
your angle is therefore in quadrant 1 and in quadrant 4.
the angle in quadrant 4 is 360 - 72.96875154 = 287.0312485 degrees.
your solution is that theta = 72.96875154 or 287.0312485.
round that to a tenth of a degree and your solution is:
theta = 73.0 or 287.0 degrees.
the following picture shows the solution graphically.
2 equations were graphed.
they are:
y = sin^2(theta)
y = 3 - 4cos(theta)
the intersection of the 2 equations on the graph is when their values are equal.
that occurs at theta = 73 and theta = 287 between 0 and 360 degrees.
the graphical solution conforming to the algebraic solution is shown below:
this graph is of the equation y = 2cos^2(theta) - 4cos(theta) + 1
the solution, in this case, is when the graph crosses the x-axis.
